Question

A thin glass rod is bent into a semi circle of radius $r$. A charge $+q$ is uniformly distributed along the upper half and a charge $-q$ is uniformly distributed along the lower half, as shown in the figure. The magnitude and direction of the electric field $\vec{E}$ produced at $P$, the centre of the circle, will be

• A. 0
• B. $\frac{q}{{\epsilon }_0{\pi }^2{r}^2}$ perpendicular to the line OP and directed downward
• C. $\frac{q}{{\epsilon }_0\pi r^2}$ perpendicular to the line OP and directed downward
• D. $\frac{q}{{\epsilon }_0\pi r^2}$ along the axis OP

Answer 1

Answer: (B)

Take $PO$ as the $x$-axis and $PA$ as the $y$-axis. Consider two elements $EF$ and $E^{\prime }{F}^{\prime }$ of width $d\theta$ at angular distance $\theta$ above and below PO, respectively.
The magnitude of the field at $P$ due to either element is
$dE=\frac{1}{4\pi {\epsilon }_0}\frac{rd\theta \times Q/(\pi r/2)}{r^2}=\frac{Q}{2{\pi }^2{\epsilon }_0{r}^2}d\theta$
Resolving the fields, we find that the components along PO sum up to zero, and hence the resultant field is along PA.
Therefore, field at $P$ due to pair of elements is $2E\sin \theta$
$E=\underset{0}{\overset{\pi /2}{\int }}2E\sin \theta$
$=2\underset{0}{\overset{\pi /2}{\int }}\frac{Q}{2{\pi }^2{\epsilon }_0{r}^2}\sin \theta d\theta =\frac{Q}{{\pi }^2{\epsilon }_0{r}^2}$